Solve sqrt{25-16}+3^4-(7-2)^2+5^0*3

$$\sqrt { 25 - 16 } + 3 ^ { 4 } - ( 7 - 2 ) ^ { 2 } + 5 ^ { 0 } \cdot 3$$

Evaluate

$$62$$

Solution Steps

Subtract 16 from 25 to get 9.

$$\sqrt{9}+3^{4}-\left(7-2\right)^{2}+5^{0}\times 3$$

Calculate the square root of 9 and get 3.

$$3+3^{4}-\left(7-2\right)^{2}+5^{0}\times 3$$

Calculate 3 to the power of 4 and get 81.

$$3+81-\left(7-2\right)^{2}+5^{0}\times 3$$

Add 3 and 81 to get 84.

$$84-\left(7-2\right)^{2}+5^{0}\times 3$$

Subtract 2 from 7 to get 5.

$$84-5^{2}+5^{0}\times 3$$

Calculate 5 to the power of 2 and get 25.

$$84-25+5^{0}\times 3$$

Subtract 25 from 84 to get 59.

$$59+5^{0}\times 3$$

Calculate 5 to the power of 0 and get 1.

$$59+1\times 3$$

Multiply 1 and 3 to get 3.

$$59+3$$

Add 59 and 3 to get 62.

$$62$$

Factor

$$2\times 31$$